What is (\partial)/(\partial x)(ln(2+sin(xy))) ?

The answer to (\partial)/(\partial x)(ln(2+sin(xy))) is (ycos(xy))/(2+sin(xy))

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(\partial)/(\partial x)(ln(2+sin(xy)))

\frac{\partial}{\partial x} (ln (2 + sin (x y)))

\frac{y cos ( x y )}{2 + sin ( x y )}

Solution steps

\frac{\partial}{\partial x} (ln (2 + sin (x y)))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 + sin ( x y )} \frac{\partial}{\partial x} (2 + sin (x y))

= \frac{1}{2 + sin ( x y )} \frac{\partial}{\partial x} (2 + sin (x y))

\frac{\partial}{\partial x} (2 + sin (x y)) = y cos (y x)

= \frac{1}{2 + sin ( x y )} y cos (y x)

Simplify

\frac{1}{2 + sin ( x y )} y cos (y x) : \frac{y cos ( x y )}{2 + sin ( x y )}

= \frac{y cos ( x y )}{2 + sin ( x y )}

\frac{\partial}{\partial z} (z x^{2})

\frac{d y}{d x} = sin (x) (cos (2 y) - cos^{2} (y))

n = 1 \sum \infty \frac{n - 1}{n !}

x \to \infty lim (\frac{e ^{5 x} + x ^{2}}{e ^{5 x} - x})

\frac{d}{d t} (2 sin^{3} (t))

What is (\partial)/(\partial x)(ln(2+sin(xy))) ?
The answer to (\partial)/(\partial x)(ln(2+sin(xy))) is (ycos(xy))/(2+sin(xy))